A sporting goods store believes the average age of its customers is 38 or less. A random sample of 40 customers was surveyed, and the average customer age was found to be 41.6 years. Assume the standard deviation for customer age is 8.0 years. Using alphaequals0.05, complete parts a and b below. a. Does the sample provide enough evidence to refute the age claim made by the sporting goods store? Determine the null and alternative hypotheses. Upper H 0: mu less than or equals 38 Upper H 1: mu greater than 38 The z-test statistic is 2.85. (Round to two decimal places as needed.) The critical z-score(s) is(are) 1.64. (Round to two decimal places as needed. Use a comma to separate answers as needed.) Because the test statistic is greater than the critical value, reject the null hypothesis. b. Determine the p-value for this test. The p-value is . 998. (Round to three decimal places as needed.)

A random sample of 40 customers was surveyed, and the average customer age was found to be 41.6 years and the standard deviation for customer age is 8.0 years.

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] <= 38 years

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 38 years

The test statistics we will use here is;

T.S. = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, X bar = sample mean = 41.6 years

[tex]\sigma[/tex] = population standard deviation = 8 years

n = sample size = 40

So, test statistics = [tex]\frac{41.6-38}{\frac{8}{\sqrt{40} } }[/tex] = 2.85

At 5% level of significance, the z table gives critical value of 1.6449. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject the null hypothesis and conclude that average age of its customers is more than 38 years.

P-value is given by, P(Z > 2.85) = 1 - P(Z <= 2.85) = 1 - 0.99781 = 0.002 .